Identify the initial value. х у -15/2 o 5 1 10 2 20 3 40 الان

Please help, il mark brainliest for the right answer

kari74 [83]

Answer:

I think that it may be 33

Because, you see 15-9=6

so you do 20-9=11

Then you do 25-9=16

add them all together to get 33 so i think that's the perimeter

I am really not sure if this is correct...i'm kinda bad at stuff like this

6 0

3 years ago

How does the value of the digit 3 in the number 63,297 compare to the value of the digit 3 in the number 60,325

Iteru [2.4K]

Answer:

Step-by-step explanation:

The three in the first number is in the thousands place and in the second number the three is in the hundreds place.

4 0

3 years ago

Read 2 more answers

Simplify: (-3x^3+2x^2+4x)-(8x^4-5x^3+6x^2-2x

sergejj [24]

Answer:

−8^4+2^3−4^2+6

Step-by-step explanation:

Distribute the - to 8x^4-5x^3+6x^2-2x:

-8x^4+5x^3-6x^2+2x

Now simplify:

-3x^3+2x^2+4x-8x^4+5x^3-6x^2+2x= −8^4+2^3−4^2+6

Hope this helps!

3 0

2 years ago

What is c? Please answer this

Black_prince [1.1K]

Answer:

C is a letter on the alphabet

Step-by-step explanation:

you havent included any pictures so i dont know what you mean

6 0

3 years ago

Read 2 more answers

A contractor is required by a county planning department to submit one, two, three, four, or five forms (depending on the nature

Westkost [7]

Answer:

(a) The value of k is $\frac{1}{15}$ .

(b) The probability that at most three forms are required is 0.40.

(c) The probability that between two and four forms (inclusive) are required is 0.60.

(d) $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ is not the pmf of y.

Step-by-step explanation:

The random variable Y is defined as the number of forms required of the next applicant.

The probability mass function is defined as:

$P(y) = \left \{ {{ky};\ for \ y=1,2,...5 \atop {0};\ otherwise} \right$

(a)

The sum of all probabilities of an event is 1.

Use this law to compute the value of k.

$\sum P(y) = 1\\k+2k+3k+4k+5k=1\\15k=1\\k=\frac{1}{15}$

Thus, the value of k is $\frac{1}{15}$ .

(b)

Compute the value of P (Y ≤ 3) as follows:

$P(Y\leq 3)=P(Y=1)+P(Y=2)+P(Y=3)\\=\frac{1}{15}+\frac{2}{15}+ \frac{3}{15}\\=\frac{1+2+3}{15}\\ =\frac{6}{15} \\=0.40$

Thus, the probability that at most three forms are required is 0.40.

(c)

Compute the value of P (2 ≤ Y ≤ 4) as follows:

$P(2\leq Y\leq 4)=P(Y=2)+P(Y=3)+P(Y=4)\\=\frac{2}{15}+\frac{3}{15}+\frac{4}{15}\\ =\frac{2+3+4}{15}\\ =\frac{9}{15} \\=0.60$

Thus, the probability that between two and four forms (inclusive) are required is 0.60.

(d)

Now, for $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ to be the pmf of Y it has to satisfy the conditions:

$P(y)=\frac{y^{2}}{50}>0;\ for\ all\ values\ of\ y \\$
$\sum P(y)=1$

Check condition 1:

$y=1:\ P(y)=\frac{y^{2}}{50}=\frac{1}{50}=0.02>0\\y=2:\ P(y)=\frac{y^{2}}{50}=\frac{4}{50}=0.08>0 \\y=3:\ P(y)=\frac{y^{2}}{50}=\frac{9}{50}=0.18>0\\y=4:\ P(y)=\frac{y^{2}}{50}=\frac{16}{50}=0.32>0 \\y=5:\ P(y)=\frac{y^{2}}{50}=\frac{25}{50}=0.50>0$

Condition 1 is fulfilled.

Check condition 2:

$\sum P(y)=0.02+0.08+0.18+0.32+0.50=1.1>1$

Condition 2 is not satisfied.

Thus, $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ is not the pmf of y.

7 0

3 years ago